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3 years ago

5

In which direction is the magnetic force acting on the charge?

1 answer:

aleksandr82 [10.1K]3 years ago

7 0

Answer:

Force on a moving charge inside magnetic field is given by

So as per right hand rule if we put our fingers of right hand along the velocity of charge and then curl it towards the magnetic field direction then thumb will show the direction of force on moving charge.

Step-by-step explanation:

If we use the same for the given condition then velocity is given out of the plane and magnetic field is downwards then as per right hand rule direction of force on the moving charge will be "Towards Right"

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Y^2+X^2=36 y=-3x+5.

lesya [120]

Answer:

The two solutions in exact form are:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$

and

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$ .

If you prefer to look at approximations just put into your calculator:

$(3.3303,-4.9909)$

and

$(-0.3303,5.9909)$ .

Step-by-step explanation:

I guess you are asked to find the solution the given system.

I'm going to use substitution.

This means I'm going to plug the second equation into the first giving me:

$(-3x+5)^2+x^2=36$ I replaced the 1st y with what the 2nd y equaled.

Before we continue solving this I'm going to expand the $(-3x+5)^2$ using the following:

$(a+b)^2=a^2+2ab+b^2$ .

$(-3x+5)^2=(-3x)^2+2(-3x)(5)+(5)^2$

$(-3x+5)^2=9x^2-30x+25$

Let's go back to the equation we had:

$(-3x+5)^2+x^2=36$

After expansion of the squared binomial we have:

$9x^2-30x+25+x^2=36$

Combine like terms (doing the $9x^2+x^2$ part:

$10x^2-30x+25=36$

Subtract 36 on both sides:

$10x^2-30x+25-36=0$

Simplify the 25-36 part:

$10x^2-30x-11=0$

Compare this to $ax^2+bx+c=0$ which is standard form for a quadratic.

We should see the following:

$a=10$

$b=-30$

$c=-11$

The formula that solves this equation for the variable $x$ is:

$x=\frac{-b \pm \sqrt{b^2-4ac}}{2a}$

Plugging in our values for $a,b, \text{ and } c$ give us:

$x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{2(10)}$

Simplify the bottom; that is 2(10)=20:

$x=\frac{30 \pm \sqrt{(-30)^2-4(10)(-11)}}{20}$

Put the inside of square root into the calculator; that is put $(-30)^2-4(10)(-11)$ in the calculator:

$x=\frac{30 \pm \sqrt{1340}}{20}$

Side notes before continuation:

Let's see if 1340 has a perfect square.

I know 1340 is divisible by 10 because it ends in 0.

1340=10(134)

134 is even so it is divisible by 2:

1340=10(2)(67)

1340=2(2)(5)(67)

1340=4(5)(67)

1340=4(335)

4 is a perfect square so we can simplify the square root part further:

$\sqrt{1340}=\sqrt{4}\sqrt{335}=2\sqrt{335}$ .

Let's go back to the solution:

$x=\frac{30 \pm \sqrt{1340}}{20}$

$x=\frac{30 \pm 2 \sqrt{335}}{20}$

Now I see all three terms contain a common factor of 2 so I'm going to divide top and bottom by 2:

$x=\frac{\frac{30}{2} \pm \frac{2 \sqrt{335}}{2}}{\frac{20}{2}}$

$x=\frac{15 \pm \sqrt{335}}{10}$

So we have these two x values:

$x=\frac{15+\sqrt{335}}{10} \text{ or } \frac{15-\sqrt{335}}{10}$

Now we just need to find the corresponding y-coordinate for each pair of points.

I'm going to use the easier equation $y=-3x+5$ .

Let's do it for the first x I mentioned:

If $x=\frac{15+\sqrt{335}}{10}$ then

$y=-3(\frac{15+\sqrt{335}}{10})+5$ .

Let's simplify:

Distribute the -3 to the terms on top:

$y=\frac{-45-3\sqrt{335}}{10}+5$

Combine the two terms; I'm going to do this by writing 5 as 50/10:

$y=\frac{-45-3\sqrt{335}+50}{10}$

Combine like terms on top; the -45+50 part:

$y=\frac{5-3\sqrt{335}}{10}$ .

So one solution point is:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$ .

Let's find the other one for the other x that we got.

If $x=\frac{15-\sqrt{335}}{10}$ then

$y=-3(\frac{15-\sqrt{335}}{10})+5$ .

Let's simplify.

Distribute the -3 on top:

$y=\frac{-45+3\sqrt{335}}{10}+5$

I'm going to write 5 as 50/10 so I can combine the terms as one fraction:

$y=\frac{-45+3\sqrt{335}+50}{5}$

Simplify the -45+50 part:

$y=\frac{5+3\sqrt{335}}{10}$ .

So the other point of intersection is:

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$ .

The two solutions in exact form are:

$(\frac{15+\sqrt{335}}{10},\frac{5-3\sqrt{335}}{10})$

and

$(\frac{15-\sqrt{335}}{10},\frac{5+3\sqrt{335}}{10})$ .

If you prefer to look at approximations just put into your calculator:

$(3.3303,-4.9909)$

and

$(-0.3303,5.9909)$ .

3 0

3 years ago